53. In the derivation of \"ransition state theory.\" the S8. Which soquence of s

Question

53. In the derivation of "ransition state theory." the S8. Which soquence of state-to-state transitions for a diatomic assumption is made that the reactants are system is required for one to observe fluorescence wth a wavelength longer than that for the excitation photon? (A) unstable relative to the products (B) in equilibrium with the products (C) in equilibrium with an activated complex (D) hard sphere molecules. Singlet 54. For the reaction H+H-HH-H+H Which point represents the transition state? (A) (B) 2(C) 3 (D) 55. If an atom has a doubly degenerate excited state that is de cm-1 above the non-degenerate ground state, the electronic partition function at any temperature (T) for this 56. The rate expression for the decomposition of a substrate (S) in the presence of an enzyme (E) is The mechanism of an enzyme catalyzed reaction of a substrate (S) to yield prodacts (P) can be written 59. As IS] bocomes very large compared to Km the apparent order for the substrase (S) is where ES and EP represent the enzyme-substrate and enzyme -product complexes, respectively. For this mechanism, the potential energy diagram (potential cnergy or enthalpy versus reaction coordinate) would be expected to have how many maxima? (A) zero() one (C) two (D) two-thirds 7. For the with this energy level diagram to operate as hv hv lase) (A) (B) two (C) thr (D) four 60. For the concerted reaction A+B-, C + D LASER the rate constants must have these relationships: E (forward) is 46 J-mol-1 and AH--27 kJmor Therefore, E, (in units of kJ-mol-) for the reverse (A) kk) and k3) must be very smal. (B) k)>k) and k3) must be large (A) 19 (B)7 (C) 46 () 73 (C) k>kcz) and ka) must be small. (D) kO>k) and k3) must be large.

Explanation / Answer

53. Option C

In activated complex theory or transition state theory an activated complex is formed from the reactants and it forms an equilibrium with the reactants. So option c is right answer.

56. As substrate concentration is too large then in the denominator, Equilibrium constant us negligible when compared to [S]. So rate is depending on initial concentration of enzyme [E]o. Thus it becomes first order. So option isB

57. Option A

58.A

59. Two intermediate compounds ES and EP are formed during the course of reaction. So two Maxima are observed.

60. Activation energy for reverse reaction = Activation energy of forward reaction + heat of reaction

Activation energy= 46+27= 73kJ-mol-1

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